Answers

2014-05-20T12:07:54+05:30
Given tanA = 1/2   and tanB = 1/3
 we know that 
tan(A+B) = (tanA + tanB)/1-tanA*tanB
tan(A+B) = (1/2+1/3)/(1-1/2*1/3)
               =(5/6)/(1-1/6)
             = (5/6)/(5/6)
              =1
tan (A+B)=tan 45                          ( since tan 45 = 1)
therefore A+B= 45
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2014-05-20T13:24:09+05:30
Given tanA = 1/2   and tanB = 1/3
We already know that 

tan(A+B) = \frac{tanA+tanB}{1-TanAtanB}

tan(A+B ) = \frac{ \frac{1}{2} + \frac{1}{3}}{1- \frac{1}{2} \frac{1}{3}}

tan(A+B ) = \frac{ \frac{5}{6}}{1- \frac{1}{6}}

tan(A+B ) = \frac{ \frac{5}{6}}{\frac{5}{6}}

tan(A+B) = 1
tan (A+B) = tan 45^o

therefore A+B = 45^o
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